Question

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.50×105 V/m . When the space is filled with dielectric, the electric field is E= 2.40×105 V/m .
What is the charge density on each surface of the dielectric?

Answer 1

Explanation:

The given data is as follows.

     Electric field between plates without dielectric, $E_{1} = 3.50 \times 10^{5} V/m$

   Electric field between the plates with dielectric, $E_{2} = 2.40 \times 10^{5} V/m$.

  Permittivity of free space, $\epsilon_{o}$ = $8.85 \times 10^{-12} C^{2}/Nm^{2}$

Now, we will determine the charge density as follows.

            $\sigma_{i} = \epsilon_{o}(E_{1} - E_{2})$

                 = $8.85 \times 10^{-12} \times (3.50 \times 10^{5} - 2.40 \times 10^{5})$

                 = $9.735 \times 10^{-7} C/m^{2}$

Thus, we can conclude that the charge density on each surface of the dielectric is $9.735 \times 10^{-7} C/m^{2}$.